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UNIT : 7
DUAL NATURE OF RADIATION AND MATTER
GIST OF THE UNIT:
Work function-The minimum amount of energy required by an electron to just escape
from the metal surface is known as work function of the metal.
One Electron Volt (1eV)-It is the kinetic energy gained by an electron when it is
accelerated through a potential difference of 1 volt.
Photon-According to Planck's quantum theory of radiation, an electromagnetic wave
travels in the form of discrete packets of energy called quanta. One quantum of light
radiation is called a photon.
The main features of photons are as follows:-
(i) A photon travels with the speed of light.
(ii) The rest mass of a photon is zero i.e., a photon cannot exist at rest.
(iii) Energy of a photon, E = hv
(iv) Momentum of a photon, p =mc
Photoelectric Effect: -The phenomenon of emission of electrons from a metallic surface
when light of appropriate frequency (above threshold frequency) is incident on it, known
as photoelectric effect.
It’s experiment was performed by Heinrich Hertz.
It was explained by Einstein and he also got Nobel Prize for this.
Photoelectric Effect experimental setup-
Factors which Effect Photoelectric Effect:-
(i) Effect of potential on photoelectric current: It is can be shown in fig. potential v/s
photoelectric current.
243
(ii) Effect of intensity of incident radiations on photoelectric current: For frequency of
radiations as constant.
(iii) Effect of frequency of the incident radiations on stopping potential: For constant
intensity.[CBSE F 2011]
The stopping potential Vo depends on(i) The frequency of incident light and (ii) the
nature of
emitter material. For a given frequency of incident light, the stopping potential is
independent
of its intensity.
eVo =(1/2)m 2
max maxv K
244
From this graph between frequency ν, stopping potential Plank's constant (h) can be
determined
Law of Photoelectric effect-
De-Broglie Hypothesis-
According to de Broglie, every moving particle is associated with a wave which controls
the particle in every respect. The wave associated with a particle is called matter wave or
de Broglie wave.
λ =
=
245
This is known as de-Broglie equation.
de-Broglie wavelength of an electron of kinetic energy K-
Davisson and Germer experiment-
This experiment proves the existence of de-Broglie waves.It establishes the wave nature
of electron particle.
Theory-A sharp diffraction is observed in the electron distribution at an accelerating
voltage of 54 V and scattering angle 50°. The maximum of intensity obtained in a
particular direction is due to constructive interference of electrons scattered from
different layers of the regularly spaced atoms of the crystal.
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FORMULAE AND SHORT CUT FORMULAE
ALL POSSIBLE FORMULA
1. Energy of a photon E =hʋ =
2. Number of photon emitted per second N =
3. Momentum of photon P = mc =
=
=
4. Equivalent mass of photon m =
=
=
5. Work function W0 = hʋ0 =
=
6. Kinetic energy of photoelectron is given by Einstein’s photoelectric
equation: Kmax
= = hʋ - W0 = h(ʋ - ʋ0) = h (
-
)
7. If V0 is the stopping potential, the maximum kinetic energy of the ejected
photoelectron, K =
=eV0
8. Kinetic energy of De-Broglie Waves K = = P
2/2m
9. Momentum of De-Broglie Waves P =√
10. Wavelength of De-Broglie Waves =
=
=
√
11. De –Broglie Wavelength of an electron beam accelerated through a
potential difference of V volts is
=
√ =
√ nm =
√ A
0
12. De –Broglie Wavelength associated with gas molecules of mass m at
temperature T kelvin is
λ =
√ K = Boltzmann constant
13. The value of hc = 12400eV A0
14. The Value of
= 1240 X 10
-9 eV m
15. Bragg’s equation for crystal diffraction is 2dSinθ = nλ, n is the order of
the spectrum
247
CONCEPT MAP
248
MIND MAP
249
Identification of important topics/concepts for slow learners
Concepts Degree of
importance
Common errors committed
Photons *** 1.Rest mass of photon 2.Unable to understand quantum nature
of photons
Photoelectric effect *** 1.Unable to apply concept of
threshold frequency or threshold
wavelength
Threshold frequency
and wavelength
** 1.Concept of threshold
frequency not clear Work function ** 1. Conversion of Joule into eV
and vice versa Experimental study
of photoelectric
effect
* 1.Concept of stopping
potential not clear.
2. Effect of intensity and frequency
on photoelectric effect.
Laws of photoelectric
Effect
**** 1.Graphical representation of
effect of intensity and frequency not
clear
Einsteins
Photoelectric
equation
*** 1.Explanation of laws of
photoelectric effect on the basis of
Einstein’s equation
de Broglie waves **** 1.Momentum of photon and
moving particle Wave nature of
Electrons
*** 1.Confusion in formulae to be
applied Davisson and Germer
Experiment
**** 1.Nature of polar graphs not
clear
2. Braggs law not clear.
Important derivation type and theoretical(knowledge based) type of questions which
covering whole unit/concepts( 3 marks &5 marks) with answers
1.. What are photons? State their important properties.
250
Ans. Photons. According to Planck’s quantum theory of radiation, an electromagnetic wave
travels in the form of discrete packets of energy called quanta. One quantum of light radiation
is called a photon. The main features of photons are as follows:
(i) A photon travels with the speed of light.
(ii) The frequency of a photon does not change as it travels from one medium to
another.
(iii) The speed of a photon changes as it travels through different media due to the
change in its wave-length.
(iv) The rest mass of a photon is zero ie a photon can not exist at rest.
(v) Energy of a photon, E = hv = hc/ʎ.
(vi) Momentum of a photon, p = mc = hv/c = h/ ʎ.
(vii) From Einstein’s mass-energy relationship, the equivalent mass m of a photon is
given by E = mc² = hv or m = hv/c².
2. Briefly describe the observations of Hertz, Hallawachs and Lenard in regard of
photoelectric effect.
Ans. Hertz’s Observations. The phenomenon of photoelectric effect was discovered by
Heinrich Hertz in 1887. While demonstrating the existence of electromagnetic waves, Hertz
found that high voltage sparks passed across the metal electrodes of the detector loop more
easily when the cathode was illuminated by ultraviolet light from an arc lamp. The ultraviolet
light falling on the metal surface caused the emission of negatively charged particles, which are
now known to be electrons, into the surrounding space and hence enhanced the high voltage
sparks.
Hallawach’s and Lenard’s Observations. During the years 1886-1902, Wilhelm
Hallwachs and Philipp Lenard investigated the phenomenon of photoelectric emission in detail.
As shown in Fig, Hallwachs connected a zinc plate to an electroscope. He allowed
ultraviolet light to fall on a zinc plate. He observed that the zinc plate became (i) unchanged if
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initially negatively charged, (ii) positively charged if initially uncharged and (iii) more
positively charged if initially positively charged. From these observations, he concluded that
some negatively charged particles were emitted by the zinc plate when exposed to ultraviolet
light.
A few years later, Lenard observed that when ultraviolet radiations are allowed to fall on
the emitter plate of an evacuated glass tube enclosing two electrodes (cathode C and anode A), a
current flows in the circuit, as shown in figure 11.2. As soon as ultraviolet radiations are
stopped, the current also stops. In 1900, Lenard argued that when ultraviolet light is incident on
the emitter plate, it causes the emission of electrons from its surface. These electrons are
attracted by the positive collector plate so that the circuit is completed and a current flows. This
current is called photoelectric current.
Hallwachs and Lenard also observed that when the frequency of the incident light was les
than a certain minimum value, called the threshold frequency, no photoelectrons were emitted at
all.
3. Describe an experimental arrangement to study photoelectric effect. Explain the effect of
(i) intensity of light on photoelectric current, (ii) potential on photoelectric current, and (iii)
frequency of incident radiation on stopping potential.
Or
Describe suitable experiments to study the laws of photoelectric emission.
Ans. Experimental study of photoelectric effect. Fig shows the experimental arrangement
used for the study of photoelectric effect. It consists of an evacuated glass/quartz tube which
encloses a photosensitive plate C and another metal plate A. The two electrodes are connected
to a high tension battery, a potential divider arrangement and a microammeter μA. When
monochromatic radiations of a sufficiently high frequency fall on the plate C, electrons are
emitted which are collected by the plate A. So a current, called photoelectric current, flows in
the outer circuit which is measured by the microammeter
μA.
(i). Effect of intensity of light on photoelectric
current. If we allow radiations of a fixed frequency to
fall on plate P and the accelerating potential difference
between the two electrodes is kept fixed, then the
photoelectric current is found to increase linearly with the
intensity of incident radiation, as shown in Fig 11.4.
Since the photoelectric current is directly proportional to
the number of photoelectrons emitted per second, it
252
implies that the number of photoelectrons emitted per second is proportional to the intensity of
incident radiation.
(ii) Effect of potential. As shown in Fig 11.5, if we keep the intensity I1 and the
frequency of incident radiation fixed and increase the positive potential (called accelerating
potential) on plate A gradually, it is found that the photoelectric current increases with the
increase in accelerating potential till a stage is reached when the photoelectric current becomes
maximum and does not increase further with the increase in the accelerating potential. This
maximum value of the photoelectric current is called the saturation current. At this stage, all the
electrons emitted by the plate C are collected by the plate A.
Now, if we apply a negative potential on plate A with respect to plate C and increase its
magnitude gradually, it is seen that the photoelectric current decreases rapidly until it becomes
zero for a certain value of negative potential on plate A. The value of the retarding potential at
which the photoelectric current becomes zero is called cut off or stopping potential for the given
frequency of the incident radiation. At the stopping potential Vo, when no photoelectrons are
emitted, the work done by stopping potential on the fastest electron must be equal to its kinetic
energy.
253
Hence K max = ½ m v²max = e Vo
If we repeat the experiment with incident radiation of the same frequency but of higher
intensity I2 and I3 (I3 > I2 > I1 ), we find that the values of saturation currents have increased
in proportion to the intensity of incident radiation, while the stopping potential is still the same.
Thus, for a given frequency of incident radiation, the stopping potential is independent of its
intensity.
(iii) Effect of frequency of incident radiation on stopping potential. To study the
effect of frequency on photoelectric effect, the intensity of incident radiation at each frequency
is adjusted in such a way that the saturation current is same each time when the plate A is at a
positive potential. The potential on the plate A is graduallyreduced to zero and then increased in
the negative direction till stopping potential is reached. The experiment is repeated with
radiations of different frequencies.
As shown in Fig the value of stopping potential increases with the frequency of incident
radiation. For frequencies v3 > v2 > v1 , the corresponding stopping potentials vary in the order
Vo3 > Vo2 > Vo1.
If we plot a graph between the frequency of incident radiation and the corresponding
stopping potential for different metals, we get straight line graphs, as shown in Fig 11.7. These
graphs reveal the following facts:
(i) The stopping potential or the maximum kinetic energy of the photoelectrons
increases linearly with the frequency of the incident radiation, but is independent of its
intensity.
(ii) There exists a certain minimum cut off frequency for which the stopping potential
is zero. The minimum value of the frequency of incident radiation below which the
photoelectric emission stops altogether is called threshold frequency.
(iii) For two different metals A and B these graphs have some slope. But the threshold
frequencies are different.
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4. State the laws of photoelectric emission.
Ans. Laws of photoelectric emission. On the basis of the experimental results on
photoelectric effect, Lenard and Millikan gave the following laws of photoelectric emission:
1. For a given photosensitive material and frequency of incident radiation, (above the
threshold frequency), the photoelectric current is directly proportional to the intensity of
light. The saturation current is directly proportional to the intensity of incident radiation.
2. For a given photosensitive material, there exists a certain minimum cut off
frequency below which no photoelectrons are emitted, howsoever high is the intensity of
incident radiation. This frequency is called threshold frequency.
3. Above the threshold frequency, the stopping potential or equivalently the
maximum kinetic energy of the photoelectrons is directly proportional to the frequency of
incident radiation, but is independent of its intensity.
4. The photoelectric emission is an instantaneous process. The time lag between the
incidence of light radiation and the emission of photoelectrons is very small, even less
than 10-9
s.
5. Explain why photoelectric effect cannot be explained on the basis of wave nature of light.
Ans. Faliure of wave theory to explain photoelectric effect. According to wave theory, light is
an electromagnetic wave consisting of electric and magnetic fields with continuous distribution
of energy over the region over which the wave extends. This theory failed to explain the
following exp. facts:
1. A higher intensity of incident radiation should liberate photoelectrons of higher
KE. But Kmax. Is found to be independent of intensity.
255
2. No matter what the frequency of incident radiation is, a light wave of sufficiently
intensity should be able to eject the electrons from the meatl surface. Thus the
wave theory fails to explain the existence of threshold frequency.
3. An electron intercepts a very small amt. of the energy of a light wave and so it
should require a finite time to escape from metal surface. But actually, the emission is
almost instantaneous.
6. Establish Einstein’s photoelectric equation. Use this equation to explain the laws of
photoelectric emission.
Ans. Einstein's theory of photoelectric effect. In 1905,Einstein explained photoelectric effect on
the basis of Planck's quantum theory according to which a light radiation travels in the form of
discrete photons. The energy of each photon is hv. Photoelectric emission is the result of
interaction of two particles — one a photon of incident radiation and the other an electron of
photosensitive metal. The energy hv of the incident photon is used up in two parts:
(a) a part of the energy of the photon is used in liberating the electron from the metal surface,
which is equal to the work function W₀ of the metal, and
(b) the remaining energy of the photon is used in imparting kinetic energy to the ejected
electron.
Therefore, Energy of the incident photon = Maximum K.E. of photoelectron + Work function
256
Or hv =½mvmax+w₀Or
Kmax =½mvmax =hv —W₀ ...(1)
If the incident photon is of threshold frequency V₀ , then its energy hv₀ is just sufficient to free
the electron from the metal surface and does not give it any kinetic energy. So hv₀ =W₀. Hence
Kmax =½mvmax =hv—hv₀ =h(v—v₀) / ...(2)
Equations (1) and (2) are called Einstein ’s photoelectric equations and can be used to explain
the laws of photoelectric effect as follows :
1. Explanation of effect of intensity. The increase of intensity means the increase in the
number of photons striking the metal surface per unit time. As each photon ejects only one
electron, so the number of ejected photoelectrons increases with the increase in intensity of
incident radiation.
2. Explanation of threshold frequency. If v <v₀ i.e., the frequency of incident radiation is less
than the threshold frequency, the kinetic energy of photoelectrons becomes negative. This has
no physical meaning. So photoelectric emission does not occur below the threshold frequency.
3. Explanation of kinetic energy. If v >v₀ , then Kmax =½mvmax v
i.e., above the threshold frequency, the maximum kinetic energy of the electrons increases
linearly with the frequency v of the incident radiation. Moreover, the increase in intensity
increases only the number of incident photons and not their energy. Hence the maximum kinetic
energy of the photoelectrons is independent of the intensity of incident radiation.
4. Explanation of time lag. Photoelectric emission is the result of an elastic collision between a
photon and an electron. Thus the absorption of energy from a photon by a free electron inside
the metal is a single event which involves transfer of energy in one lump instead of the
continuous absorption of energy as in the wave theory of light. Hence there is no time lag
between the incidence of a photon and the emission of a photoelectron.
7. Draw a graph showing the variation of stopping potential with the frequency of incident
radiation in relation to photoelectric effect.
(a) What does the slope of this graph represent ?
(b) How can the value of Planck's constant be determined from this graph ?
257
(c) How can the value of work function of the material be determined from this graph.
Ans. Determination of Planck's constant and work function. According to Einstein's
photoelectron equation, the maximum K.E. of a photoelectron is given by
Kmax =hv —W₀
If V₀ is the stopping potential, then
Kmax = eV₀
Therefore, eV₀ = hv —W₀
V₀=(h/e)v —W₀/e
It follows from the above equation that V₀ versus v is a straight line, as shown in Figure
(a) Clearly, slope of V₀ —v graph =h/e
(b) To determine the slope, take two points A and B on the straight line graph. Then
m = tan θ = AC/BC = h/e
h = e × AB/BC=e × slope of V₀ —v graph
Thus, the Planck's constant h can be determined.
Therefore, moreover, the intercept on vertical axis = —W₀/e
W₀ = e × Magnitude of the intercept on vertical axis.
In this way, the work function W₀ can be determined
258
8 . What considerations led de-broglie to suggest that material particles can also have wave
properties? What are de-Broglie waves.
Ans. Dual nature of matter : de-Broglie waves. In 1924, the French physicist Louis Victor de—
Broglie (pronounced as de Broy) put forward the bold hypothesis that material particles in
motion should display wave-like properties. He reasoned that nature was symmetrical and two
basic physical entities-matter and energy, must have symmetrical character. If radiation shows
dual nature, so should matter. 50 the particles like electrons, protons, neutrons, etc. should
exhibit wave nature when in motion. The waves associated with material particles in motion are
called matter or de Broglie waves and their wavelength is called de Broglie wavelength.
9 . Show that the de Broglie wavelength λ of electrons of energy K is given by the relation :
λ
√
Ans. de-Broglie wavelength of an electron of kinetic energy K. Consider an electron mass m
and charge e. Let v be the velocity attained by the electron when it is accelerated from rest
attain kinetic energy K. Then . 1‘
½mv²= K
Or v √
√
Hence the de-Broglie wavelength of the electron is
λ=h/mv=
√
or λ=
√
Q10. Show that the de-Broglie wavelength λ of electrons accelerated through a potential
difference of V
volts can expressed as
λ=
√
√ A
ans. de-Broglie wavelength of an electron accelerated through a potential difference I Consider
an electron of mass m and charge e. Let v be the velocity acquired by it when it is accelerated
from rest through a potential difference of V volts. Then
259
K.E. gained by the electron =½mv²
Work done on electron = eV
K.E. gained by electron = Work done on electron
½mv²=eV
Or v √
√
Hence de-Broglie wavelength of electrons is
λ=h/mv=
√
or λ=
√
but h=6.63×10⁻³⁴ Js,
m=9.1×10⁻³¹Kg
e=1.6×10⁻¹⁹
λ ⁻ ⁴
√
√
λ=
√ A
Q11 .Describe Davisson and Germer experiment to establish the wave nature of electrons.
Describe a
labelled diagram of the apparatus used.
Ans. Davisson and Germer experiment. In this experiment, as shown in Figure, the electrons
emitted by the hot filament of an electron gun are accelerated by applying a suitable potential
difference V between the cathode and anode. The fine collimated beam of electrons from the
electron gun is directed against the face Of Ni crystal. The crystal is capable of rotation about
an axis perpendicular to the plane Of paper. The electrons, scattered in different directions by
the atoms of Ni crystal, are received by a movable detector which is just an electron collector.
260
Thus we measure scattered electron intensity as a function Of the scattering angle Ф, the angle
between the incidence and the scattered electron beam. The experiment is repeated for different
accelerating potentials V.
Figures show the results of Davisson and Germer experiment, when the accelerating voltage
was varied from 44 V to 68 V. Clearly, there is a strong peak corresponding to a sharp
diffraction maximum in the electron distribution at an accelerating voltage of 54 V and
scattering angle 50°. The maximum of intensity obtained in a particular direction is due to
constructive interference of electrons scattered from different layers of the regularly spaced
atoms of the crystal.
From Figure, the glancing angle θ is given by
θ+Ф+θ=180
θ=90
Ф=90
The interatomic separation for Ni crystal is
d= 0.914 A
For first order (n =1) diffraction maximum, the Bragg’s law is
261
2d sin θ = λ
λ=2×0.914× sin65
From de-Broglie hypothesis, the wavelength associated with an electron beam accelerated
through 54 V must be
λ=
=
√ =
√ =1.66
The experimentally measured wavelength is close to that estimated from de—BrogIie
hypothesis. This proves the existence of de—Broglie waves.
Previous 8 years AISSCE questions
1 mark questions
1. Write the expression for the de-Broglie wavelength associated with a charged particle having
charge q and mass m when it is accelerated by a potential. (All
India 2013)
Ans: The charged particle as a mass m and charge q. The kinetic energy of the particle is equal
to the work done on it by the electric field
E = qV
mv
2 = qV => P = √
2. Define intensity of radiation in photon picture of light. (All India
2012)
Ans: The photon incident on unit area of a surface in unit time.
3. Define the term stopping potential in relation to photoelectric effect. (All
India 2011)
Ans: The value of negative potential of anode at which photoelectric current in the circuit
reduces to zero is called stopping potential.
4. For a given photosensitive material and with a source of constant frequency of incident
radiation. How does the photocurrent vary with the intensity of incident light?
(All India 2011)
Ans: (i) I3 >I2 > I1
photocurrent
I3
I2
I1
-V0 collector plate potential
262
5. The stopping potential in an experiment on photoelectric effect is 1.5 V. What is the
maximum kinetic energy of the photoelectrons emitted?
(All India 2009)
Ans: KEmax = eV0 => KEmax = 1.5 eV
6. The maximum kinetic energy of a photoelectron is 3 eV. What is its stopping potential?
(All India
2009)
Ans: KEmax = 3eV => 3eV = eV0 => V0 = 3V
7. The stopping potential in an experiment on photoelectric effect is 2 V. What is the maximum
kinetic energy of the photoelectrons emitted? (All
India 2009)
Ans: KEmax = eV0 = e(2V) = 2 eV
8. Ultraviolet radiations of different frequencies ν1, ν2 are incident on two photo sensitive
materials having work functions W1 and W2 (W1>W2) respectively. The kinetic energy of
emitted electrons is same in both the cases. Which one of the two radiations will be at the
higher frequency? (All India 2007)
Ans: KE = hν – W
(KE)1 = (KE)2
hν1 – W1 = hν2 – W2 here W1> W2 => ν1 > ν2
9. A proton and an electron have same kinetic energy. Which one has greater de-Broglie
wavelength and why? (All
India 2012)
Ans: λ =
=
√ , λe > λp
10. Two lines A and B in the given shows the variation of de-Broglie wavelength λ versus
√ ,
where V is the accelerating potential difference for two particles carrying the same charge.
Which one of the two represents a particle of smaller mass?
(All India 2008)
λ
B
A
√
263
Ans: λ =
=
√ The slope of λ versus
√ graph will be inversely proportional to the square
root of mass of the particles. Now slope of B is Greater it represents mass is smaller.
11. The de-Broglie wavelength associated with an electron accelerated through a potential
difference V is λ. What will be its wavelength when the accelerating potential is increased to
4 V? (All India 2006)
Ans:
√ ̇ for V = 4 V
√ ̇ = 6.135 ̇
12. Name the experiment which establishes the wave nature of a particle.
(All India 2006C)
Ans: Davission-Germer experiment
2 marks questions 13. Write Einstein’s photoelectric equation. State clearly the three salient features observed in
photoelectric effect which can be explained on the basis of above equation.
(All India 2010)
Ans: Einstein’s equation for photoelectric effect
hν = KEmax + W0
KEmax = h (ν – ν0)
KEmax = h(
)
Three salient features
(i) Threshold frequency for KEmax ≥ 0 => ν ≥ ν0 Photo electric effect takes place
(ii) KEmax of photoelectron ν > ν0 => KEmax α (ν - ν0)
(iii)Effect of intensity of incident light : The number of photon incident per unit timr per
unit area increases with the increase of incident light
14. Draw suitable graphs to show the variation of photoelectric current with collector plate
potential for (i) a fixed frequency but different intensities I1>I2>I3. (ii) a fixed intensity but
different frequencies ν1>ν2> ν3.
(All India 2010)
Ans: (i) I3 >I2 > I1
photocurrent
I3
I2
I1
-V0 collector plate potential
264
(ii) ν03 > ν02 > ν01
photocurrent
ν3 ν2 ν1
collector plate potential
-V03 –V02 –V01
15. Write Einstein’s photoelectric equation relating the maximum kinetic energy of the emitted
electron to the frequency of the radiation incident on a photosensitive surface. State clearly
the basic elementary process involved in photoelectric effect.
(All India 2009C)
Ans: Einstein’s equation for photoelectric effect
hν = KEmax + W0
KEmax = h (ν – ν0)
Basic elementary process
(i) Threshold frequency for KEmax ≥ 0 => ν ≥ ν0 Photo electric effect takes place
(ii) KEmax of photoelectron ν > ν0 => KEmax α (ν - ν0)
(iii)Effect of intensity of incident light : The number of photon incident per unit timr per
unit area increases with the increase of incident light
16. Write Einstein’s photoelectric equation in terms of the stopping potential and the threshold
frequency for a given photosensitive material. Draw a plot showing the variation of stopping
potential versus the frequency of incident radiation. V0
(All India 2008)
Ans: hν = KEmax + W0
KEmax = hν – W0
But W0 = hν0
KEmax = hν – hν0 ν
eV0 = h (ν – ν0) V01
V0 =
(ν – ν0)
Slope the graph gives ratio of value of plank constant (h) V02
and electronic charge (e)
265
17. An α-particle and a proton are accelerated from rest by the same potential. Find the ratio of
their de-Broglie wavelengths. (All
India 2010)
Ans: λ =
=
√
= √
=
√
18. The two lines marked A and B in the given figure show a plot of de-Broglie wavelength λ
versus
√ , where V is the accelerating potential for two nuclei
(i)What does the
slope of the lines represent? (ii) Identify which of the lines corresponded to these nuclei.
(All India 2008)
λ
B
A
√
Ans: (i) The slope of the line represent
√
(ii)
carry same charge as they have same atomic number √ α
√
The lighter mass represented by line greater slope i. e., A and similarly
by line B.
19. Crystal diffraction experiments can be performed either by using electrons accelerated
through appropriate voltage or by using X-rays. If the wavelength of these probes (electrons
or X-ray) is 1 ̇ estimate which of the two has greater energy.
(All India 2009)
Ans: The de-Broglie matter wave equation is given by
λ =
=
√ =>
----(i)
for X-ray photon of same wavelength E’ =
-----(ii)
=
=
< 1 => K< E’ Energy possess by X-ray is more than electron.
20. For what kinetic energy a neutron will the associated de-Broglie wavelength be 1.32 x 10-10
m? (All India
2008)
Ans: λ =
=
√ =>
=> K = 7.5 x 10
-21 J
266
21. For what kinetic energy a neutron will the associated de-Broglie wavelength be 1.32 x 10-10
m?(Given mass of neutron 1.675 x 10-27
kg)
(All India 2008)
Ans: : λ =
=
√ =>
=> K = 7.5 x 10
-21 J
HOTS questions with answers (2 marks &3 marks)--05 questions
1. A source of light is placed at a distance of 50cm from a photocell and the cut-off
potential is found to be V0.If the distance between the light source and photocell is made
25 cm ,What will be the new cut-off potential? Justify your answer
Ans: The stopping potential is still V0.As the distance is decreased from 50cm to
25cm,the intensity of light becomes four times the original intensity.But the stopping
potential is independent of the intensity.
2. Two metal X and Y when illuminated with appropriate radiation emit photoelectron. The
work function of X is higher than that of Y .Which metal will have higher value of
threshold frequency and why?
Ans: The work function of metal a metal is given by w = h0 ,where 0 is the threshold
frequency.as the work function of metal X is higher ,therefore it will also have higher
value of threshold frequency.
3. Two photon of energy 2.5 eV each are incident simultaneously on a surface with work
function 4.3 eV.Will photoelectric emission take place? Why or Why not?
Ans :-One photon can eject one photoelectron from the surface of a metal provided its
energy is not less than the work function of metal.Since each each photon has energy less
than tha work function of given surface,Hence photoelectric emission is not possible.
4. It is harder to remove an electron from Cu than Na.Which metal has greater work
function? Which has greater threshold wavelength?
Ans:The work function of sodium is less than that of the copper ;therefore,it is easier to
remove an electron from sodium than from copper;
As work function :- w = h0 so threshold frequency:- 0 =
;It show that work
function of copper is
more than that of sodium and threshold wavelength is less
5 .A beam of light consists of four wavelengths 4000Ǻ, 4800Ǻ, 6000Ǻ & 7000Ǻ, each of
intensity 1.5mW/m2. The beam falls normally on an area 10
-4m
2 of a clean metallic surface of
work function 1.9eV.Assuming no loss of kinetic energy, calculate the number of
photoelectrons emitted per second.
Ans :E1 = 3.1eV, E2 = 2.58eV, E3 = 2.06eV, E4 = 1.77eV
Only the first three wave lengths can emit photo electrons.
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Number of photo electrons emitted per second = IA ( 1/E1+1/E2+1/E3 )
= 1.12 x 1012
.
( Hint – convert eV into joule before substitution )
6..Find the ratio of wavelength of a 10 k eV photon to that of a 10 keV electron.
2
Ans: 10 ( Hint: λphoton = 1.24 A0, λelectron = 0.1227 A
0 )
7.A parallel beam of light is incident normally on a plane surface absorbing 40% of the light
and reflecting the rest. If the incident beam carries 10W of power, find the force exerted by it on
the surface.
Ans : 5.33 x 10-8
N
8.No photoelectrons are emitted from a surface, if the radiation is above 5000 Ǻ. With an
unknown wavelength, the stopping potential is 3V. Find the wave length.
Ans : 2262Ǻ
9.Illuminating the surface of a certain metal alternately with light of wave lengths0.35μm and
0.54μm, it was found that the corresponding maximum velocities of photoelectrons have a ratio
2. Find the work function of that metal.
Ans: 5.64eV
Value based questions with answers -( 5questions)
1. In a multistoried building, once a fire broke out at midnight due to electrical short circuit.
Ravi along with others rushed to the spot, informed the fire service and put off the fire.
But by that time a huge amount of damage had already been done. Ravi being Secretary
of the building decided to fix fire alarms (using photo cell) in all the floors
Q – (i) What values were shown by Ravi in this situation?
Ans . Concern for society, social responsibility, application of knowledge.
Q-(ii) A human eye can perceive a minimum light intensity is about 10-10
Wm-2 .Calculate the number of photons of wavelength 5.6 x 10-7 m/s that
must enter the pupil of area 10-4 m2 for vision ?
Ans. Energy falling on area per sec =10-10 x 10-4 = nhc / λ
Therefore n= λ10-10 x 10-4 / hc = 3 x 104.photons/sec
2. Davisson and Germer's actual objective was to study the surface of a piece of nickel by
directing a beam of electrons at the surface and observing how many electrons bounced
off at various angles. Though the experiment was conducted in a vacuum chamber. Air
entered the chamber, producing an oxide film on the nickel surface. When they started
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the experiment again and the electrons hit the surface, they were scattered by atoms
which
originated from crystal planes inside the nickel crystal. Davisson and Germer's accidental
discovery of the diffraction of electrons was the first direct evidence confirming de
Broglie's hypothesis that particles can have wave properties as well.
(i)What can we infer about the values shown by Davisson and Germer?
Ans: Perseverance, not giving up and patience.
(ii) Write the expression to find the wavelength of an electron when accelerated through a
potential difference of Vvolts.
Ans : =
√ =
√ nm =
√ A
0
3. Ravi, while returning home from office late night, saw a person jumping into a house
.Suspecting him to be a thief, he slowly followed him. He saw an alarm fixed at the entry
of the house. Immediately to alarm the people at the house, he used his laser torch to
activate the alarm, as it works with the visible light. The circuit became complete
and siren started working and people at the house got up and caught the thief. They
thanked Ravi for his action.
a) What moral value can you see in Ravi?
Ans: Answer: Presence of mind/ Social responsibility/ attitude to help others.
b) The photoelectric cut off voltage in a certain experiment is 1.5V. What is the maximum
kinetic energy of the photoelectrons emitted?
Ans: 1.5 eV
4 A seminar was conducted in an auditorium in Chennai. It was a big auditorium and some seats
at the back were vacant. The coordinator had a problem in making the order for lunch, as he
was unable to estimate the number of people present in the auditorium. Vikram , associate of
the coordinator , saw a digital counter working on the principle of photo electric effect,at every
entry and exit gate of the auditorium. The people were allowed to enter through the entry gate
and leave the hall by the exit gate. He also noticed that all the counters were in good condition.
He immediately noted the display of every counter and gave the coordinator the exact number
Of people in the auditorium.
a) What moral values can you see in Vikram?
Ans: Answer: a)Proper application of knowledge/ concern for his boss/ presence
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b) What is photo electric effect?.
Q5.. A function was arranged in the school auditorium. The auditorium has the capacity of 400
students. When entry started students entered in groups and counting becomes a great
problem. Then science students took responsibility at the gate. All the students entered the
hall one by one. This helped them to maintain discipline and counting became easy with the
help of a device used by these students.
(i) What value is displayed by science students?
(ii) Name the device which is based on application of photoelectric effect.
Ans: (i) Sense of responsibility.
(ii) Photo cell.
Numerical Problems based on important concepts/frequently asking in AISSCE
1. Write the expression for the de-Broglie wavelength associated with a charged particle having
charge q and mass m when it is accelerated by a potential.
Ans: The charged particle as a mass m and charge q. The kinetic energy of the particle is
equal to the work done on it by the electric field
E = qV
mv
2 = qV => P = √
2. The stopping potential in an experiment on photoelectric effect is 1.5 V. What is the
maximum kinetic energy of the photoelectrons emitted?
Ans: KEmax = eV0 => KEmax = 1.5 eV
3. The maximum kinetic energy of a photoelectron is 3 eV. What is its stopping potential?
Ans: KEmax = 3eV => 3eV = eV0 => V0 = 3V
4. Ultraviolet radiations of different frequencies ν1, ν2 are incident on two photo sensitive
materials having work functions W1 and W2 (W1>W2) respectively. The kinetic energy of
emitted electrons is same in both the cases. Which one of the two radiations will be at the
higher frequency?
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Ans: KE = hν – W
(KE)1 = (KE)2
hν1 – W1 = hν2 – W2 here W1> W2 => ν1 > ν2
5. A proton and an electron have same kinetic energy. Which one has greater de-Broglie
wavelength and why?
Ans: λ =
=
√ , λe > λp
6. Two lines A and B in the given shows the variation of de-Broglie wavelength λ versus
√ ,
where V is the accelerating potential difference for two particles carrying the same charge.
Which one of the two represents a particle of smaller mass?
λ
B
A
√
Ans: λ =
=
√ The slope of λ versus
√ graph will be inversely proportional to the square
root of mass of the particles. Now slope of B is Greater it represents mass is smaller.
7. The de-Broglie wavelength associated with an electron accelerated through a potential
difference V is λ. What will be its wavelength when the accelerating potential is increased to
4 V?
Ans:
√ ̇ for V = 4 V
√ ̇ = 6.135 ̇
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8. An α-particle and a proton are accelerated from rest by the same potential. Find the ratio
of their de-Broglie wavelengths.
Ans: λ =
=
√
= √
=
√
9. For what kinetic energy a neutron will the associated de-Broglie wavelength be 1.32 x 10-10
m?
Ans: λ =
=
√ =>
=> K = 7.5 x 10
-21 J
10. For what kinetic energy a neutron will the associated de-Broglie wavelength be 1.32 x 10-10
m?(Given mass of neutron 1.675 x 10-27
kg)
Ans: : λ =
=
√ =>
=> K = 7.5 x 10
-21 J
